Of 97. A. 4.610436. Of 199. A. 5.838272. Of 179. A. Of 981. A. 9.936261. 29791 Proceed with fractions as in extracting the square root. 9. Find the cube root of 13. A.. Of Of 148333. A. 77. Of 73,01394. A. 134. A. 334. 273 13934. A. 24 Of 2 2 3 4 6 4 4 3 10. Find the cube root of §. A. .8549879. Of 19 2570773 Of. A. 20. A. .4562903. From these illustrations and examples, we derive the rule, 1. HAVING POinted off, SUBTRACT FROM THE HIGHEST PERIOD, THE GREATEST CUBE CONTAINED IN IT, PLACE THE ROOT IN THE QUOTIENT, AND, TO THE REMAINDER, BRING DOWN THE NEXT PERIOD FOR A DIVIDEND. II. SQUARE The root alreaDY FOUND, (UNDERSTANDING A CYPHER AT THE RIGHT,) AND MULTIPLY IT BY 3 FOR A DIVISOR. DIVIDE THE DIVIDEND BY THE DIVISOR FOR THE NEXT FIGURE OF THE ROOT. III. MULTIPLY THE DIVISOR BY THE QUOTIENT, MULTIPLY 3 TIMES THE SQUARE OF THE QUOTIENT BY THE PART OF THE ROOT PREVIOUSLY FOUND, FINALLY, CUBE THE QUOTIENT, AND ADD THESE THREE RESULTS TOGETHER FOR A SUBTRAHEND. IV. SUBTRACT the subtrahend from THE DIVIDEND, TO THE REMAINDER, BRING DOWN THE NEXT Period for a new dividend, and so PROCEED. NOTE. The pupil's attention should be particularly called to the effect of understanding a cypher in 11 and 111. If he does not fully understand this, he will be liable to fall into error. We have preferred the above rule, to the one usually given, because it keeps the principles of the operation before the mind, and, indeed, obliges the pupil to act upon them at almost every step. Perhaps it is not as easy as the other, in the beginning, but it will be found far more useful, and in fact, easier in the end. The proof is by Involution. ARITHMETICAL PROGRESSION. § CII. The natural series of numbers, 1, 2, 3, 4, 5, &c., consists of numbers increasing by a continual addition of 1 to the number preceding. The series, 1, 3, 5, 7, 9, &c., increases by a continual addition of 2. The series, 15, 12, 9, 6, &c., decreases by a continual subtraction of 3. Any rank, or series of numbers, consisting of more than two terms, increasing or decreasing, like the above, by a common difference, is called an ARITHMETICAL SERIES OF PROGRESSION. The numbers forming the series, are called TERMS. The first and last terms, are called the EXTREMES, the others, the MEANS. When the series increases, or is formed by a continual addition of the common difference, it is called an ASCENDING SERIES. When it decreases, or is formed by a continual subtraction, of the common difference, it is called a DESCENDING SERIES. 1. How many strokes do the clocks in Venice, which go from 1 to 24 o'clock, strike in the course of a day? The answer might be found by addition. But by attending to the following, an easier mode may be discovered. Suppose another clock made to strike downward from 24 to 1. The two clocks would strike equal numbers of blows in a day. The order of their strik. ing would be as follows: 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24 24,23,22,21,20,19,18,17,16,15,14,13,12,11,10, 9, 8, 7, 6, 5, 4, 3, 2, 1 25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25 Thus, at every hour, the two clocks together would strike the same number of blows, viz. 25, and, in the course of the day, both would strike 24×25-600 blows. One would strike half this number= 300 Ans. It will be seen that 25=24+1=the sum of the extremes, Hence, the extremes and number of terms being given, to find the sum of all the terms, MULTIPLY THE SUM OF THE EXTREMES BY THE NUMBER OF TERMS, AND HALF THE PRODUCT WILL BE THE ANSWER. 2. The first term of a series is 1, the last term 29, and the num. ber of terms 14. What is the sum of the series? A. 210. 3. 1st. term, 2, last term, 51, number of terms, 18. Required the sum of the series, A. 477. 4. Find the sum of the natural terms 1, 2, 3, &c. to 10,000. Ans. 50,005,000. 5. A man travelled from Hartford, going 3 miles the first day, and increasing each day by an equal excess. His 12th day's journey was 58 miles. What was the daily increase, and the distance he travelled from Hartford ? As he travelled 12 days, he increased his journey 11 times, by an equal addition. 58, then, is 11 times the daily increase, more than 3. Therefore 58-3=55÷11–5 daily increase. By last rule, 366 mls, distance from H. Hence, the extremes and number of terms being given, to find the common difference, DIVIDE THE DIFFERENCE OF THE EXTREMES BY THE NUMBER OF TERMS LESS 1. 6. Extremes 3 and 19; number of terms 9. dif. Required the com.. in value, from the 7. Extremes 4 and 56; number of terms 14. dif. A man had 15 houses, increasing equally first, worth $700, to the 15th worth $3,500. rence in value between the first and second ? 9. The ages of 5 persons were in arithmetical progression, the youngest being 15 yrs. old, and the com. dif. 2. What was the eldest one's age? It was evidently 4 times the com. dif. more than that of the youngest. Hence, 15+4X2=23 Ans. Hence, the least term, the number of terms, and the common difference being given, to find the greatest term, TO THE LEAST term, add the COMMON Difference, multiplied by THE NUMBER OF TERMS LESS 1. 4. 100 stones are a yard apart in a straight line, and the first one a yard from a basket. How far must a man go, to gather them, one by one, and return with them singly to the basket? Ans. 5 mls. 5 fur. 36 rds. 2 yds. NOTE. The cases of Arithmetical Progression are numerous, and of too little practical utility to warrant us in devoting to them much space. Any three of the following terms being given, the other two may be found. 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms. GEOMETRICAL PROGRESSION. § CIII. The series 1, 3, 9, 27, 81, &c., consists of numbers, each of which is 3 times the preceding. The series 64, 32, 16, 8, 4, &c., consists of numbers, each of which is half the preceding. Any rank or series of numbers, of more than two terms, increasing, like the above, by a common multiplier, or decreasing by a common divisor, is called a GEOMETRICAL SERIES, or PROGRESSION. The common multiplier or divisor is called the RATIO. The numbers which form the series are called TERMS. The distinction between increasing or ascending, and decreasing or descending series, is made as in Arithmetical Progression. Any three of the five following terms being given, the other two may be found. 1. The first term. 2. The last term. 3. The num. ber of terms. 4. The ratio. 5. The sum of all the terms. 1. A man bought 5 sheep, giving $1 for the first; $3 for the sec. ond; $9 for the third, and so on, in geometrical progression. What did he give for the whole? Write the whole series, thus, 1, 3, 9, 27, 81 Multiply by the ratio, 3, thus, 3, 9, 27, 81, 243 We here obtain a new series, having 4 terms like the last. Let the first series be subtracted from the second, and all the similar terms vanish, leaving 1, the 1st. term of the 1st. series, to be sub. tracted from 243, the last term of the last. 243-1-242. Now as the last series was 3 times the first, and we have subtracted once the first, the remainder, 242, must be twice the first series. HENCE, 242÷2121 sum of 1st, series, Ans. The same result would have been obtained, if we had multiplied only the last term, and taken the first from the product, since the intermediate terms vanish, in the subtraction. Hence, the rule, the extremes and ratio being given, to find the sum of all the terms. MULTIPLY THE GREATEST TERM BY THE RATIO, FROM THE PRODUCT SUBTRACT THE LEAST TERM, AND DIVIDE THE REMAINDER BY THE RATIO, LESS 1. 2. Given the first term, 1; the last term, 2,187; and the ratio, 3: required the sum of the series. A. 3,280 3. Extremes, 1 and 65,536; ratio 4: required the sum of the series. A. 87,381 4. Extremes 1,024 and 59,049: required as above. A. 175,099. 5. What is the sum of the series 16, 4, 1, 1, rb, 64, and so on, to an infinite extent ? A. 211. NOTE. The last term is evidently 0. 6. A man had a debt to pay as follows: the first month $3; the second, $6; the third, $12, and so on, in a twofold ratio. In 12 months the debt was paid. What was the last payment? The second payment is found by multiplying the first by the ratio, 2, once; the third, by multiplying by 2, twice; and so on, where it will be seen that the ratio is always used as a multiplier, a single time less than the number of terms. Hence, the 12th. term the first term the ratio eleven times successively, or, in other words, the eleventh power of the ratio, thus, 3×2×2×2×2×2×2×2×2×2×2×2=3×211=6,144 Ans. Hence, the first term, ratio, and number of terms being given, to find the last term, MULTIPLY THE FIRST TERM BY THAT POWER OF THE RATIO, WHOSE INDEX IS 1 LESS THAN THE NUMBER OF TERMS. NOTE. In involving the ratio, it will be seen that the process may often be abridged by multiplying together two powers already obtained. Thus, the 3d power the 2d power the 5th power, &c. 7. If the first term is 2, the ratio, 2, and the number of terms, 13, what is the last term? A. 8,192. 8, Find the 12th. term of a series, whose 1st. term is 3, and ratio, 3. A. 531,441. 9. First term, 1, ratio 2; required the 23d. term. A. 4,194,304. 10. A man bought a horse, giving 1 ct. for the first nail in his shoes, three for the second, and so on; there were 32 nails; what cost the horse. ? A. 9,265,100,944,259.20. 11. A man works for a farmer 40 years, receiving 1 kernel of corn for the first year, 10 for the second, and so on; what do his wages amount to, allowing 1,000 kernels to a pt., and supposing corn worth 50 cts. per bu.? A. $8,680,555,555,555,555,555,555,555,555,555,555.555555 12. A young man agreed to work eleven years with a farmer, on condition of receiving the produce of one wheat corn, the first year; the produce of that quantity, sowed the second year; and so on, to the end of the time. How much wheat was there due for his service, and what would it come to at $1 per bu., allowing the yearly increase to have been tenfold, and 7,680 corns to make a pt.? A. 226,056) bu, at $226,056.125 NOTE. The cases of Geometrical Progression are numerous. Ten different ones may be stated, with two requisitions in each case, making, in fact, 20 different cases. The same is true in Arithmetical Progression. It would require much space to give a clear understanding of all these cases, and as they are of little practical importance, we pass them by. The terms Arithmetical and Geometrical, Are used without regard to their proper signification, (viz. belonging to Arithinetic and to Geometry,) but only to distinguish these different kinds of Progression. ANNUITIES. CIV. 1. A pension is allowed a man and his heirs, forever, of $600. He is willing to dispose of this pension at a fair price. What ought he to receive for it, allowing money to draw 6 per ct. in terest ? It is evident he ought to receive a sum which would produce an annual interest, equal to the pension. $600 then is the interest, and the principal is acquired at 6 pr. ct. Hence, (§ LXXXI.) $600÷.06= $10,000 Ans. A sum of money, payable periodically, for a certain length of time, or forever, is called an ANNUITY. An annuity, in the proper sense of the word is a sum, payable annually. Payments, made at greater or less periods, are, however called annuities. To annuities 'belong pensions, salaries, rents, &c. When an annuity remains unpaid after it is due, it is said to be in arrears. The sum of the annuities in arrears, with the interest on each, is called the AMOUNT. The sum, which ought to be paid for an annuity yet to come, is called the PRESENT WORTH. From the above example, it is evident that, to find the present worth of an annuity to continue forever, we must DIVIDE THE ANNUITY BY THE RATE PER CENT. 2. What should be paid for a perpetual annuity of $40, discount. ing at 5 pr. ct.? Ans. 800. 3. What is an estate worth, which brings in $7,500 a year allowing 6 pr. ct.? Ans. $125,000. ANNUITIES AT SIMPLE INTEREST. 1. If a rent of $600 be in arrears 5 years, what will be due at 4 per cent.? On the rent of every year but the 5th., interest is due; for the 4th. year, one year's interest, for the 3d, two years' interest, and so on, in Arithmetical Progression. The question falls, therefore, under § CII. $24 the com. dif. 600+24×4=696 largest term. 696+600 X5÷2 $3,240 Ans. Hence, to find the amount of an annuity in arrears, FIND THE SUM OF AN ARITHMETICAL PROGRESSION, OF WHICH THE ANNUITY IS THE FIRST TERM, THE NUMBER OF YEARS, THE NUMBER OF TERMS, AND THE ANNUAL INTEREST, THE RATIO; AND IT WILL BE THE AMOUNT REQUIRED. |